ENGINEERING MECHANICS STATIC

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Hubert Stone
5 years ago
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Trusses Simple trusses The basic element of a truss is the triangle, three bars joined by pins at

The term rigid is used to mean noncollapsible and also to mean that deformation of the members due

On the other hand, four or more bars pin-jointed to form a polygon of as many sides constitute a

b rigid, or stable, by adding a diagonal bar joining A and D or B and C and thereby forming two

1 Trusses Simple trusses The basic element of a truss is the triangle, three bars joined by pins at their ends, fig. a below, constitutes a rigid frame. The term rigid is used to mean noncollapsible and also to mean that deformation of the members due to induced internal strains is negligible. On the other hand, four or more bars pin-jointed to form a polygon of as many sides constitute a nonrigid frame. We can make the nonrigid frame in fig. b rigid, or stable, by adding a diagonal bar joining A and D or B and C and thereby forming two triangles. We can extend the structure by adding additional units of two end-connected bars, such as DE and CE or AF and DF, fig c, which are pinned to two fixed joints. In this way the entire structure will remain rigid. Asst. lecturer Sarmed A. Salih Page 42

2 Structures built from a basic triangle in the manner described are known as simple trusses. To design a truss we must first determine the forces in the various members and then select appropriate sizes and structural shapes to withstand the forces. Several assumptions are made in the force analysis of simple trusses. First, we assume all members to be two-force members. A two-force member is one in equilibrium under the action of two forces only. Each member of a truss is he two points of application of force. The two forces are applied at the ends of the member and are necessarily equal, opposite, and collinear for equilibrium. The member may be in tension or compression, as shown in fig. below. When we represent the equilibrium of a portion of a two-force member, the tension T or compression C acting on the cut section is the same for all sections. We assume here that the weight of the member is small compared with the force it supports. If it is not, or if we must account for the small effect of the weight, we can replace the weight W of the member by two forces, each W/2 if the member is uniform, with one force acting at each end of the member, these forces, in effect, are treated as loads externally applied to the pin connections. Accounting for the weight of a member in this way gives the correct result for the average tension or compression along the member but will not account for the effect of bending of the member. Asst. lecturer Sarmed A. Salih Page 43

3 EX 14 Sol Compute the force in each member of the loaded cantilever truss. The first step will be to compute the forces at D and E from the free-body diagram of the truss as a whole. The equations of equilibrium give. E = 0 T = 0 T = 80 kn 80 cos 30 -E X = 0 E X = 69.3 kn 80 sin 30 + E Y = 0 E Y = 10 kn Equilibrium for Joint A gives. AB sin = 0 AB = 34.6 kn T AC 34.6 cos 60 = 0 AC = kn C Where T stands for tension and C stands for compression Equilibrium for Joint B gives. BC sin sin 60 = 0 BC = 34.6 kn C BD cos 60 = 0 BD = 34.6 kn T Asst. lecturer Sarmed A. Salih Page 44

Equilibrium for Joint C gives. CD sin 60-34.6 sin 60-20 = 0 CD = 57.7 kn T CE 17.32-34.6 cos 60-57.7 cos 60 = 0 :. CE = 63.5 kn C Equilibrium for Joint E gives. DE sin 60-10 = 0 DE = 11.

4 Equilibrium for Joint C gives. CD sin sin = 0 CD = 57.7 kn T CE cos cos 60 = 0 :. CE = 63.5 kn C Equilibrium for Joint E gives. DE sin = 0 DE = kn C For check we us the cos 60 = 0 EX 15 Determine the force in each member of the loaded truss. Equilibrium for Joint C gives. CD sin 30-3 = 0 CD = 6 kn C 6 cos 30 BC = 0 BC = 5.2 kn T BC 30 CD 3 kn Asst. lecturer Sarmed A. Salih Page 45

5 Equilibrium for Joint D gives. ENGINEERING MECHANICS STATIC BD DE sin 60 6 sin 60 = 0 DE = 6 kn C DE kN BD - 6 cos 60-6 cos 60 = 0 BD = 6 kn T Equilibrium for Joint B gives. AB sin30 6 = 0 AB = 12 kn T AB BE kn BE cos 30 = 0 BE = 5.2 kn C 6 kn Equilibrium for Joint E gives. AE 6 sin30 - AE = 0 AE = 3 kn T E 5.2 kn 30 6kN Asst. lecturer Sarmed A. Salih Page 46

6 H.W 1. Determine the force in each member of the simple equilateral truss. [J. L. Merim (4-1)] Ans. AB = 736 N T, AC = 368 N T, BC = 736 N C 2. Determine the force in each member of the loaded truss. All triangles are equilateral. [J. L. Merim (4-9)] Ans. AB = kn C, AE = 8.66 kn T BC = 5 kn C, BD = 5.2 kn C, BE = 4 kn C CD = 7.5 kn T, DE = 6.35 kn T 3. A snow load transfers the forces shown to the upper joints of a Pratt roof truss. Neglect any horizontal reactions at the supports and solve for the forces in all members. [J. L. Merim (4-17)] Ans. AB = DE = BC = CD = 3.35 kn C AH = EF = 3 kn T, BH = DF = 1 kn C CF = CH = kn T, FG = GH = 2 kn T Asst. lecturer Sarmed A. Salih Page 47

Plane Trusses Trusses

TRUSSES Plane Trusses Trusses- It is a system of uniform bars or members (of various circular section, angle section, channel section etc.) joined together at their ends by riveting or welding and constructed